48. (a) Use V2y2 = V1y2
- 2*g*(change in y) and set the second velocity, at the top, equal to
zero. Find the first velocity.
(b) Find the total time by using
change in y = V1y *t - (1/2)*g*t2.
Set change y = 0 = V1y *t - (1/2)*g*t2
= t*[ V1y - (1/2)*g*t ] and
solve V1y - (1/2)*g*t = 0
for the total time.

Note you can also get the total time by getting the
time to arrive at the top of the path by using V2y
= V1y - g*t and setting the second
velocity = 0 and solving for t. The total time of flight is twice this
value !!
(c) the magnitude and direction of the acceleration are both constant (
do not change) over the entire motion .

51. (a) The total time is 8.5 seconds, so
the time to get to the top (where the velocity is zero) is 4.25 seconds.
You can find the initial velocity using :
V2y = 0 = V1y -
0.379g*t and plugging t = 4.25 seconds. Find V1y
. Then use V2y2 = 0 = V1y2 -
2*0.379g*(change in y) and find change in y.
(b) see part (a)
(c) See figure 2.28. The section of the graph you are interested
in is between 0 and 3 seconds. In the current problem you replace
3 seconds by 8.5 seconds and g by 0.379g.

53. (a) Use V2y
= V1y - g*t
and change in y = V1y *t -
(1/2)*g*t2 for the two times given. Remember to
add
40.0 m to change in y. Note the first velocity is +5.00 m/s.
(b) Use change in y = - 40.0 m = V1y
*t - (1/2)*g*t2. and solve the
quadratic equation for t. Note the first velocity is +5.00
m/s.
(c) Use V2y = V1y
- g*t and the time t from the previous part. You can also
use
V2y2 = V1y2 - 2*g*(change in
y), where change in y = - 40.0 m, in order to get the second
velocity. Note take the negative square root. The first velocity
is +5.00 m/s.
(d) Use V2y2 = V1y2
- 2*g*(change in y) and set the second velocity, at the top, equal to
zero. Find change in y. Remember to add 40.0 m to your answer.
(e) See figure 2.28 as a guide.

(b) change in y = V1y *t - (1/2)*g*t2.
Note the initial ( first) velocity is negative.
(c) Use V2y2 = V1y2 -
2*g*(change in y). Note: change in y = -10.0 m. When you solve for
the speed |V2y |, take the positive square root since you are
finding the speed. If you wanted the second velocity V2y you
would take the negative square root, since down is negative.

59. (a) From #48, we know the total time of
flight = 2* V1y / g, where V1y is the
initial velocity.
Solve for the initial velocity of the faster stone and then the initial
velocity of slower stone. Then use
time of flight = 2* V1y / g again to get the time of
flight of the slower stone.

(b) Use V2y2 = V1y2
- 2*g*(change in y) and set the second velocity, at the top, equal to
zero. Note if we are dealing with the slower stone, the change in y = H.
Solve for the initial velocity of the slow stone in terms of H. Then
find the initial velocity of the faster stone in terms of H by
using the relationship between the initial velocity of the fast
stone and the initial velocity of the slow stone.

60. In this problem you will encounter the
square root of 2.
(a) Let the first coconut (A) be dropped from H and the
second coconut (B) dropped from H/2.
VA2 = 2gH and VB2
= 2g(H/2) are the squares of the speeds when each hits the ground.
Take the ratio, then square root to get the relationship between the two
speeds at impact with ground. Note |VA | =
V. Write the B coconut speed in terms of V.
(b) |VB | = gT and |VA | = g*time. Find the
time of the A coconut in terms of T using the relationships you found in
part (a). Your answer will involve a square root of 2.

73. SOLVE A VERTICAL FREE FALL
EQUATION FOR TIME OF FLIGHT OF EGG. THEN SUBSTITUTE INTO PROFESSOR'S
UNIFORM HORIZONTAL MOTION EQUATION.
In particular, we can set the downward to be the positive y direction.
In this case,
change in y = V1y *t + (1/2)*gt2. Since the
egg is dropped, V 1y = 0. Solve for
the time t. Note the change in y is positive but is less than 46.0
m since the egg hits the head of a 1.80 m tall professor. Thus you
must subtract 1.8 m from 46 m to get change in y. The plug time t into
the formula for the horizontal displacement of the professor: x =
V*t, where V is the horizontally directed l speed of professor as she
moves to the right.

75. (a) Use V2y = V1y
- gt and find the the requested time t given V2y = + 20
m/s.
(b) Repeat for V2y = - 20 m/s
.
(c) V2y = - 40 m/s
(d) V2y = V1y - gt . Set the second velocity equal
to ZERO and find t.
(e) See figure 2.28. The section of the graph you are interested
in is between 0 and 3 seconds.

79. To begin this problem find the
height and velocity the rocket goes before shutting off the
engine.
change in y = V1y *t + (1/2)*(2.50 m/s2)*t2.
and V2y = V1y + (2.50 m/s2
) t, where the first velocity is ZERO since the rocket
begins from rest. t = 20.0 seconds.
(a) Use
VBy2 = VAy2 - 2*g*(change in
y). Set the speed at point B, the highest point, equal to ZERO.
Find the change in y and add it to the value of change of y
you got when the engine was on. Note: The VAy = V2y
you found above.
(b) At the highest point the velocity is zero but the acceleration is not
zero. (c)The time can be found in two pieces: 1. while
the engine is on and 2. while it rocket is rising and falling in free
fall
(engine shut off) .
1. The first time is 20.0 seconds
2. You know the total change in y from shut off to when it hits
the ground. It equals the negative of the vertical
displacement between launch and shut off. You calculated the
vertical displacement between launch and shut off at the beginning the
problem above: change in y = (1/2)*(2.50 m/s2)*t2,
where t = 20.0 s. Solve the following quadratic equation for t',
the time the rocket is in free fall. -change in y = V2y
*t' - (1/2)gt'2 .
The total time is 20.0 s + t'.
You can get the velocity when it hits the ground this way: V3y
= V2y - gt' .
There are other ways to solve this problem without a quadratic
equation which will be discussed in class.

83. Let t1 = time to hit the
bottom of cliff and t2 = time the sound takes to get from
bottom of cliff to your ear at the top.
Let H be the unknown cliff height. We know H = (1/2)*g*t12.
Also t2 = H/(330 m/s) and t1 = 10.0 s
- t2 . Substitute and solve for
H = (1/2)*g*( 10.0 - H/330)2. Multiply out the
right hand side and solve for H. Solve a quadratic equation for
H. Clearly if you neglected the speed of sound, you would have
overestimated the cliff height.

61. See example 2.12

64. Hint: If no wind is blowing, the
trip is simply 2*L/V, where L = 2,000 mi and V = 600 mi/h. If the
wind is blowing west to east, relative to the ground the plane has
larger speed than on the way to Chicago, and relative to the
ground the plane has a smaller speed from Chicago to SF. Thus on the way
to Chicago, the plane takes less time and on the way back the plane
takes more time.

CH. 3 Intro to Projectile Motion

USE EQUATIONS 3.12, 3.13, 3.14 AND 3.15
PAGE 77. AND SEE FIGURE 3.11 WHICH WE REVIEWED IN CLASS.
We assume up is the positive direction, such that the acceleration is
straight down and has component ay = -g. THERE IS NO
ACCELERATION IN THE X DIRECTION. THAT IS WHY THE X-COMPONENT OF VELOCITY
IS CONSTANT.

12. See example 3.3. THIS JUST
LIKE THE SIMULATION LAB. Referring to the equations mentioned above, we
see for the x-motion,
x = Vo*cos0*t = Vo*t since the launch angle is
zero. Thus Vx = Vo.
Y-MOTION:
y = -(1/2)g*t2 since sin0 = 0.
Vy = -g*t since sin0 = 0.

(a) You are given the time t = 4.00 s. Plug this time into the
y-motion equation for the vertical displacement to get
y = -20 m/s*t - (1/2)*g*t2. We are assuming
y = 0 at the initial level of the balloon. Take the absolute value
of the answer to get the initial height H of the balloon: H = |y|
(b) x = (15 m/s)*t
(c) This is a little tricky. The distance of the rock from the balloon
is given by the Pythagorean Theorem. You need to get the x displacement
of the rock from the balloon first which is given by x = (15
m/s)*t, where you plug in t = 4.0 seconds.
Now you need to get the y component of displacement of the rock from the
balloon. That simply is the difference between the height you computed
in part (a) and the vertical distance moved by the balloon in 4.0
seconds: 20.0 m/s*t - H.
The distance D between the rock and balloon is given by D2 =
x2 + (20 m/s*t - H)2. Take the square root of both
sides to get D.

56. Study examples 3.4. and 3.5. We
will review this in class. Vx will be constant for the entire
trip so the plot Vx vs t will be a straight horizontal
line (zero slope) . The plot of Vy vs t will look like
that of figure 2.28 (b), where the slope = -g.